find the sum of the infinite series (2/9)^2-(2/9)^3+(2/9)^4-(2/9)^5+ ...
(2/9)^2-(2/9)^3+(2/9)^4-(2/9)^5+
This is like a geometric series with the first terms missing
1 -1(2/9)^1 +1(2/9)^2-1(2/9)^3+(2/9)^4-(2/9)^5+
so if we find the sum of that series and subtract [1 -1(2/9)^1], we have it
so
g = 1
r = -2/9
in Sn = g (1-r^n)/(1-r)
n is infinite and |r| < 1
so Sn =g/(1-r)
Sn = 1/(1-2/9) = 1.286
so we want
1.286 -[1-2/9]
.508
Whoa - typo
Sn = 1/(1+2/9) because r = -2/9
Sn = .818
so we want
.818 -[1-2/9]
.040
i need help
To find the sum of an infinite geometric series, we can use the formula:
S = a / (1 - r)
where S is the sum of the series, a is the first term, and r is the common ratio.
In this case, the first term a is (2/9)^2 = 4/81, and the common ratio r is (2/9)^(-1) = 9/2.
Now, we can substitute these values into the formula and calculate the sum:
S = (4/81) / (1 - 9/2)
To simplify the expression, we can multiply the numerator and denominator by the reciprocal of the denominator:
S = (4/81) * (2/2 - 9) = (4/81) * (-7/2) = -28/162 = -14/81
Therefore, the sum of the given infinite series is -14/81.